Algebra Examples

$\tan (2 θ) + \tan (θ) = 0$

Step 1

Simplify each term.

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Step 1.1

Apply the tangent double-angle identity.

$\frac{2 \tan (θ)}{1 - \tan^{2} (θ)} + \tan (θ) = 0$

Step 1.2

Simplify the denominator.

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Step 1.2.1

Rewrite $1$ as $1^{2}$ .

$\frac{2 \tan (θ)}{1^{2} - \tan^{2} (θ)} + \tan (θ) = 0$

Step 1.2.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = \tan (θ)$ .

$\frac{2 \tan (θ)}{(1 + \tan (θ)) (1 - \tan (θ))} + \tan (θ) = 0$

Step 2

Factor $\tan (θ)$ out of $\frac{2 \tan (θ)}{(1 + \tan (θ)) (1 - \tan (θ))} + \tan (θ)$ .

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Step 2.1

Factor $\tan (θ)$ out of $\frac{2 \tan (θ)}{(1 + \tan (θ)) (1 - \tan (θ))}$ .

$\tan (θ) \frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} + \tan (θ) = 0$

Step 2.2

Raise $\tan (θ)$ to the power of $1$ .

$\tan (θ) \frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} + \tan (θ) = 0$

Step 2.3

Factor $\tan (θ)$ out of $\tan^{1} (θ)$ .

$\tan (θ) \frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} + \tan (θ) \cdot 1 = 0$

Step 2.4

Factor $\tan (θ)$ out of $\tan (θ) \frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} + \tan (θ) \cdot 1$ .

$\tan (θ) (\frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} + 1) = 0$

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\tan (θ) = 0$

$\frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} + 1 = 0$

Step 4

Set $\tan (θ)$ equal to $0$ and solve for $θ$ .

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Step 4.1

Set $\tan (θ)$ equal to $0$ .

$\tan (θ) = 0$

Step 4.2

Solve $\tan (θ) = 0$ for $θ$ .

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Step 4.2.1

Take the inverse tangent of both sides of the equation to extract $θ$ from inside the tangent.

$θ = \arctan (0)$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

The exact value of $\arctan (0)$ is $0$ .

$θ = 0$

Step 4.2.3

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$θ = π + 0$

Step 4.2.4

Add $π$ and $0$ .

$θ = π$

Step 4.2.5

Find the period of $\tan (θ)$ .

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Step 4.2.5.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 4.2.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 4.2.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 4.2.5.4

Divide $π$ by $1$ .

$π$

Step 4.2.6

The period of the $\tan (θ)$ function is $π$ so values will repeat every $π$ radians in both directions.

$θ = π n, π + π n$ , for any integer $n$

Step 5

Set $\frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} + 1$ equal to $0$ and solve for $θ$ .

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Step 5.1

Set $\frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} + 1$ equal to $0$ .

$\frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} + 1 = 0$

Step 5.2

Solve $\frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} + 1 = 0$ for $θ$ .

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Step 5.2.1

Find the LCD of the terms in the equation.

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Step 5.2.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$(1 + \tan (θ)) (1 - \tan (θ)), 1, 1$

Step 5.2.1.2

The LCM of one and any expression is the expression.

$(1 + \tan (θ)) (1 - \tan (θ))$

Step 5.2.2

Multiply each term in $\frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} + 1 = 0$ by $(1 + \tan (θ)) (1 - \tan (θ))$ to eliminate the fractions.

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Step 5.2.2.1

Multiply each term in $\frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} + 1 = 0$ by $(1 + \tan (θ)) (1 - \tan (θ))$ .

$\frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} ((1 + \tan (θ)) (1 - \tan (θ))) + 1 ((1 + \tan (θ)) (1 - \tan (θ))) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Simplify each term.

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Step 5.2.2.2.1.1

Cancel the common factor of $(1 + \tan (θ)) (1 - \tan (θ))$ .

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Step 5.2.2.2.1.1.1

Cancel the common factor.

$\frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} ((1 + \tan (θ)) (1 - \tan (θ))) + 1 ((1 + \tan (θ)) (1 - \tan (θ))) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2.1.1.2

Rewrite the expression.

$2 + 1 ((1 + \tan (θ)) (1 - \tan (θ))) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2.1.2

Multiply $(1 + \tan (θ)) (1 - \tan (θ))$ by $1$ .

$2 + (1 + \tan (θ)) (1 - \tan (θ)) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2.1.3

Expand $(1 + \tan (θ)) (1 - \tan (θ))$ using the FOIL Method.

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Step 5.2.2.2.1.3.1

Apply the distributive property.

$2 + 1 (1 - \tan (θ)) + \tan (θ) (1 - \tan (θ)) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2.1.3.2

Apply the distributive property.

$2 + 1 \cdot 1 + 1 (- \tan (θ)) + \tan (θ) (1 - \tan (θ)) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2.1.3.3

Apply the distributive property.

$2 + 1 \cdot 1 + 1 (- \tan (θ)) + \tan (θ) \cdot 1 + \tan (θ) (- \tan (θ)) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2.1.4

Simplify and combine like terms.

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Step 5.2.2.2.1.4.1

Simplify each term.

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Step 5.2.2.2.1.4.1.1

Multiply $1$ by $1$ .

$2 + 1 + 1 (- \tan (θ)) + \tan (θ) \cdot 1 + \tan (θ) (- \tan (θ)) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2.1.4.1.2

Multiply $- \tan (θ)$ by $1$ .

$2 + 1 - \tan (θ) + \tan (θ) \cdot 1 + \tan (θ) (- \tan (θ)) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2.1.4.1.3

Multiply $\tan (θ)$ by $1$ .

$2 + 1 - \tan (θ) + \tan (θ) + \tan (θ) (- \tan (θ)) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2.1.4.1.4

Rewrite using the commutative property of multiplication.

$2 + 1 - \tan (θ) + \tan (θ) - \tan (θ) \tan (θ) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2.1.4.1.5

Multiply $\tan (θ)$ by $\tan (θ)$ by adding the exponents.

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Step 5.2.2.2.1.4.1.5.1

Move $\tan (θ)$ .

$2 + 1 - \tan (θ) + \tan (θ) - (\tan (θ) \tan (θ)) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2.1.4.1.5.2

Multiply $\tan (θ)$ by $\tan (θ)$ .

$2 + 1 - \tan (θ) + \tan (θ) - \tan^{2} (θ) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2.1.4.2

Add $- \tan (θ)$ and $\tan (θ)$ .

$2 + 1 + 0 - \tan^{2} (θ) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2.1.4.3

Add $1$ and $0$ .

$2 + 1 - \tan^{2} (θ) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.2.2

Add $2$ and $1$ .

$3 - \tan^{2} (θ) = 0 ((1 + \tan (θ)) (1 - \tan (θ)))$

Step 5.2.2.3

Simplify the right side.

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Step 5.2.2.3.1

Expand $(1 + \tan (θ)) (1 - \tan (θ))$ using the FOIL Method.

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Step 5.2.2.3.1.1

Apply the distributive property.

$3 - \tan^{2} (θ) = 0 (1 (1 - \tan (θ)) + \tan (θ) (1 - \tan (θ)))$

Step 5.2.2.3.1.2

Apply the distributive property.

$3 - \tan^{2} (θ) = 0 (1 \cdot 1 + 1 (- \tan (θ)) + \tan (θ) (1 - \tan (θ)))$

Step 5.2.2.3.1.3

Apply the distributive property.

$3 - \tan^{2} (θ) = 0 (1 \cdot 1 + 1 (- \tan (θ)) + \tan (θ) \cdot 1 + \tan (θ) (- \tan (θ)))$

Step 5.2.2.3.2

Simplify and combine like terms.

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Step 5.2.2.3.2.1

Simplify each term.

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Step 5.2.2.3.2.1.1

Multiply $1$ by $1$ .

$3 - \tan^{2} (θ) = 0 (1 + 1 (- \tan (θ)) + \tan (θ) \cdot 1 + \tan (θ) (- \tan (θ)))$

Step 5.2.2.3.2.1.2

Multiply $- \tan (θ)$ by $1$ .

$3 - \tan^{2} (θ) = 0 (1 - \tan (θ) + \tan (θ) \cdot 1 + \tan (θ) (- \tan (θ)))$

Step 5.2.2.3.2.1.3

Multiply $\tan (θ)$ by $1$ .

$3 - \tan^{2} (θ) = 0 (1 - \tan (θ) + \tan (θ) + \tan (θ) (- \tan (θ)))$

Step 5.2.2.3.2.1.4

Rewrite using the commutative property of multiplication.

$3 - \tan^{2} (θ) = 0 (1 - \tan (θ) + \tan (θ) - \tan (θ) \tan (θ))$

Step 5.2.2.3.2.1.5

Multiply $\tan (θ)$ by $\tan (θ)$ by adding the exponents.

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Step 5.2.2.3.2.1.5.1

Move $\tan (θ)$ .

$3 - \tan^{2} (θ) = 0 (1 - \tan (θ) + \tan (θ) - (\tan (θ) \tan (θ)))$

Step 5.2.2.3.2.1.5.2

Multiply $\tan (θ)$ by $\tan (θ)$ .

$3 - \tan^{2} (θ) = 0 (1 - \tan (θ) + \tan (θ) - \tan^{2} (θ))$

Step 5.2.2.3.2.2

Add $- \tan (θ)$ and $\tan (θ)$ .

$3 - \tan^{2} (θ) = 0 (1 + 0 - \tan^{2} (θ))$

Step 5.2.2.3.2.3

Add $1$ and $0$ .

$3 - \tan^{2} (θ) = 0 (1 - \tan^{2} (θ))$

Step 5.2.2.3.3

Multiply $0$ by $1 - \tan^{2} (θ)$ .

$3 - \tan^{2} (θ) = 0$

Step 5.2.3

Solve the equation.

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Step 5.2.3.1

Subtract $3$ from both sides of the equation.

$- \tan^{2} (θ) = - 3$

Step 5.2.3.2

Divide each term in $- \tan^{2} (θ) = - 3$ by $- 1$ and simplify.

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Step 5.2.3.2.1

Divide each term in $- \tan^{2} (θ) = - 3$ by $- 1$ .

$\frac{- \tan^{2} (θ)}{- 1} = \frac{- 3}{- 1}$

Step 5.2.3.2.2

Simplify the left side.

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Step 5.2.3.2.2.1

Dividing two negative values results in a positive value.

$\frac{\tan^{2} (θ)}{1} = \frac{- 3}{- 1}$

Step 5.2.3.2.2.2

Divide $\tan^{2} (θ)$ by $1$ .

$\tan^{2} (θ) = \frac{- 3}{- 1}$

Step 5.2.3.2.3

Simplify the right side.

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Step 5.2.3.2.3.1

Divide $- 3$ by $- 1$ .

$\tan^{2} (θ) = 3$

Step 5.2.3.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\tan (θ) = \pm \sqrt{3}$

Step 5.2.3.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.2.3.4.1

First, use the positive value of the $\pm$ to find the first solution.

$\tan (θ) = \sqrt{3}$

Step 5.2.3.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$\tan (θ) = - \sqrt{3}$

Step 5.2.3.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$\tan (θ) = \sqrt{3}, - \sqrt{3}$

Step 5.2.4

Set up each of the solutions to solve for $θ$ .

$\tan (θ) = \sqrt{3}$

$\tan (θ) = - \sqrt{3}$

Step 5.2.5

Solve for $θ$ in $\tan (θ) = \sqrt{3}$ .

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Step 5.2.5.1

Take the inverse tangent of both sides of the equation to extract $θ$ from inside the tangent.

$θ = \arctan (\sqrt{3})$

Step 5.2.5.2

Simplify the right side.

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Step 5.2.5.2.1

The exact value of $\arctan (\sqrt{3})$ is $\frac{π}{3}$ .

$θ = \frac{π}{3}$

Step 5.2.5.3

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$θ = π + \frac{π}{3}$

Step 5.2.5.4

Simplify $π + \frac{π}{3}$ .

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Step 5.2.5.4.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$θ = π \cdot \frac{3}{3} + \frac{π}{3}$

Step 5.2.5.4.2

Combine fractions.

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Step 5.2.5.4.2.1

Combine $π$ and $\frac{3}{3}$ .

$θ = \frac{π \cdot 3}{3} + \frac{π}{3}$

Step 5.2.5.4.2.2

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 3 + π}{3}$

Step 5.2.5.4.3

Simplify the numerator.

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Step 5.2.5.4.3.1

Move $3$ to the left of $π$ .

$θ = \frac{3 \cdot π + π}{3}$

Step 5.2.5.4.3.2

Add $3 π$ and $π$ .

$θ = \frac{4 π}{3}$

Step 5.2.5.5

Find the period of $\tan (θ)$ .

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Step 5.2.5.5.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 5.2.5.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 5.2.5.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 5.2.5.5.4

Divide $π$ by $1$ .

$π$

Step 5.2.5.6

The period of the $\tan (θ)$ function is $π$ so values will repeat every $π$ radians in both directions.

$θ = \frac{π}{3} + π n, \frac{4 π}{3} + π n$ , for any integer $n$

Step 5.2.6

Solve for $θ$ in $\tan (θ) = - \sqrt{3}$ .

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Step 5.2.6.1

Take the inverse tangent of both sides of the equation to extract $θ$ from inside the tangent.

$θ = \arctan (- \sqrt{3})$

Step 5.2.6.2

Simplify the right side.

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Step 5.2.6.2.1

The exact value of $\arctan (- \sqrt{3})$ is $- \frac{π}{3}$ .

$θ = - \frac{π}{3}$

Step 5.2.6.3

The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the third quadrant.

$θ = - \frac{π}{3} - π$

Step 5.2.6.4

Simplify the expression to find the second solution.

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Step 5.2.6.4.1

Add $2 π$ to $- \frac{π}{3} - π$ .

$θ = - \frac{π}{3} - π + 2 π$

Step 5.2.6.4.2

The resulting angle of $\frac{2 π}{3}$ is positive and coterminal with $- \frac{π}{3} - π$ .

$θ = \frac{2 π}{3}$

Step 5.2.6.5

Find the period of $\tan (θ)$ .

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Step 5.2.6.5.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 5.2.6.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 5.2.6.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 5.2.6.5.4

Divide $π$ by $1$ .

$π$

Step 5.2.6.6

Add $π$ to every negative angle to get positive angles.

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Step 5.2.6.6.1

Add $π$ to $- \frac{π}{3}$ to find the positive angle.

$- \frac{π}{3} + π$

Step 5.2.6.6.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$π \cdot \frac{3}{3} - \frac{π}{3}$

Step 5.2.6.6.3

Combine fractions.

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Step 5.2.6.6.3.1

Combine $π$ and $\frac{3}{3}$ .

$\frac{π \cdot 3}{3} - \frac{π}{3}$

Step 5.2.6.6.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 3 - π}{3}$

Step 5.2.6.6.4

Simplify the numerator.

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Step 5.2.6.6.4.1

Move $3$ to the left of $π$ .

$\frac{3 \cdot π - π}{3}$

Step 5.2.6.6.4.2

Subtract $π$ from $3 π$ .

$\frac{2 π}{3}$

Step 5.2.6.6.5

List the new angles.

$θ = \frac{2 π}{3}$

Step 5.2.6.7

The period of the $\tan (θ)$ function is $π$ so values will repeat every $π$ radians in both directions.

$θ = \frac{2 π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer $n$

Step 5.2.7

List all of the solutions.

$θ = \frac{π}{3} + π n, \frac{4 π}{3} + π n, \frac{2 π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer $n$

Step 5.2.8

Consolidate the solutions.

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Step 5.2.8.1

Consolidate $\frac{π}{3} + π n$ and $\frac{4 π}{3} + π n$ to $\frac{π}{3} + π n$ .

$θ = \frac{π}{3} + π n, \frac{2 π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer $n$

Step 5.2.8.2

Consolidate $\frac{2 π}{3} + π n$ and $\frac{2 π}{3} + π n$ to $\frac{2 π}{3} + π n$ .

$θ = \frac{π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer $n$

Step 6

The final solution is all the values that make $\tan (θ) (\frac{2}{(1 + \tan (θ)) (1 - \tan (θ))} + 1) = 0$ true.

$θ = π n, π + π n, \frac{π}{3} + π n, \frac{2 π}{3} + π n$ , for any integer $n$

Step 7

Consolidate the answers.

$θ = \frac{π n}{3}$ , for any integer $n$